Hi David,
Yeah sorry, that was all of the top of the head, so I apologize if anything was incorrect.
I am glad you got it working though..

Best,
Karl


On May 27, 2011, at 3:50 PM, David McGlone wrote:

On Fri, 2011-05-27 at 14:48 -0500, Karl DeSaulniers wrote:
Hi David,
Try this..

if(!isset($_POST['ErrorCode'])) {
        die('You did not enter an error code!');
} else {
        $errorcode = $_POST['ErrorCode'];
        $sql = mysql_query("SELECT * FROM Sheet1 WHERE errorcode='".
$errorcode."' ");
        if($sql && mysql_numrows($sql) > 0) {
                $row = mysql_fetch_array($sql);
                for($i=0; $i<mysql_numrows($sql); $i++) {
                        echo ("Error Code: ".$row[$i]['errorcode']."<br />");
                        echo ("Error Definition: ".$row[$i]['definition']."<br 
/>");
                        echo ("<br />");
                }
        } else {
                die('No Error Code found in database');
        }
}

Have not tested this, and there may be a better "short-hand" way of
writing this, but hope it gets you on your way...

Thank you Karl, after a minute or two, or three ;-) I had it working,
but I had to change in the first line !isset to empty. For some reason
it wasn't working with !isset. and I had to
change: .$row[$i]['errorcode']. to .$row['errorcode']. because it wasn't
allowing for more than 1 digit in the error code number, but it was
displaying the correct error definition
and
 .$row[$i]['definition']. to .$row['definition']. had to be changed
because it was displaying the error code and not the definition.

Now the only problem I've got to figure out is altering it so it doesn't
exclude row[0].

Blessings,
David M.



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Karl DeSaulniers
Design Drumm
http://designdrumm.com


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